Question

When a polynomial P(x) is divided by (x - 2) and (x-3), remainders are 3 & 2 respectively. What is the remainder when the same polynomial is divided by (x-2)(x-3)?​

Answer 1

Answer:

According to factor theorem

p(x)=xq(x)+1

P(x)=(x−2)q

′

(x)+3

P(x)=(x−3)q

′′

(x)+2

So, At x=0,P(0)=1

At x=2,P(2)=3

At x=3,P(3)=2

Now when the polynomial P(x) is divided by (x−2)(x−3)x the remainder must have the degree less than 3 . that is the remainder will be of the form ax

2

+bx+c

⟹P(x)=x(x−2)(x−3)q

′′′′

(x)+ax

2

+bx+c

So, P(0)=1=a(0)

2

+b(0)+c⟹c=1

Step-by-step explanation:

Answer 2

Concept Introduction: Polynomials are a very basic concept in Mathematics.

Given:

We have been Given:

$p(x)$

is the main equation.

Quotient are

$(x - 2) \: (x - 3) \: and \: x$

Remainders are

$3 \: 2 \: and \: 1$

To Find:

We have to Find: What is the remainder when the same polynomial is divided by

$(x - 2)(x - 3)x$

Solution:

According to the problem, By Factor theorem

$p(x) = xq(x) + 1 \\ p(x - 2) = (x - 2)q(x) + 3 \\ p(x - 3) = (x - 3)q(x) + 2$

so, At,

$x = 0 \: p(0) = 1 \\ x = 2 \: p(2) = 3 \\ x = 3 \: p(3) = 2$

Now, when the Polynomial

$p(x)$

is divided by

$(x - 2)(x - 3)x$

the remainder must have the degree less than three. Therefore the remainder will be of the form

$a {x}^{2} + bx + c = 0$

therefore

$p(0) = 1 = a {(0)}^{2} + (0)x + c$

therefore,

$c = 1$

Final Answer: The remainder is

$1$

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